# Box dimension as the critical value of the fractal content

Let $$M \subseteq \mathbb{R}^n$$ be bounded and $$N_{\epsilon}(M)$$ the minimum number of 'squares' of side $$\epsilon$$ with center in M necessary to cover $$M$$. The box dimension of M is then defined as $$\dim_B \; M = \lim\limits_{\epsilon \to 0+} \frac{\log \; N_{\epsilon}(M)}{\log \; \epsilon^{-1}}$$. The following proposition shows that the s-dimensional fractal content $$N_{\epsilon}(M) \epsilon^s$$ presents a $$0-\infty$$-behaviour similar to the s-dimensional Hausdorff measure:

Proposition:

1.$$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = \infty$$ if $$s < dim_B M$$

2.$$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = 0$$ if $$s > dim_B M$$

In argument 1, I show the existence of an element $$s_0$$ where the above behaviour happens. That is, I prove the proposition replacing $$dim_B M$$ by $$s_0$$.

The difference with the Hausdorff dimension definition is that in that setting we defined the dimension as the jumping value in the observed in the $$0-\infty$$-behaviour. Here, I need to show that $$s_0 = dim_B M$$.

Until now I only have an informal argument for the case $$dim_B M \notin \{0,\infty\}$$. So:

Question

How can I formally show that $$s_0 = dim_B M$$?

Argument 1

For $$t > s \ge 0$$ write $$N_{\epsilon}(M) \epsilon^t = \epsilon^{t-s} N_{\epsilon}(M) \epsilon^s$$. If $$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s < \infty$$ then $$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^t = 0$$. On the other hand, if $$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^t < \infty$$ and $$\neq 0$$ then $$\lim\limits_{\epsilon \to 0+} N_{\epsilon}(M) \epsilon^s = \infty$$. So we can confirm a $$0-\infty$$-behaviour as shown in the diagram:

References:

Falconer, Fractal Geometry: Mathematical Foundations and Applications, 2nd edition, page 46. Elgar, Measure, Topology and Fractal Geometry, 2nd edition, page 213.

• Do you explicitly assume that the limit $\dim_B \; M = \lim_{\epsilon \to 0+} \frac{\log \; N_{\epsilon}(M)}{\log \; \epsilon^{-1}}$ exists? The lower box dimension may be strictly less than the upper box dimension. See also Falconer's book on p. 42 (2.8). – Skeeve Apr 3 at 19:37
• @Skeeve From my understanding of Falconer, yes, $dim_B \; M$ is assumed to exist and from its limit definition one should prove that it is the critical value for the fractal content. – Javier Apr 3 at 19:42
• By the way, your Argument 1 shows only that there exist $\underline s$ and $\overline s$ such that if $s> \overline s$ then $N_\epsilon \epsilon^{s} \to 0$ as $\epsilon \to 0$, and if $s < \underline s$ then $N_\epsilon \epsilon^s \to \infty$ as $\epsilon\to0$. But it is not evident that $\underline s = \overline s$. The point is that your Argument 1 does not seem to use that the limit in the definition of $\dim_B M$ exists. – Skeeve Apr 3 at 21:00

## 1 Answer

Since $$M$$ is a bounded subset of $$\mathbb R^n$$ it automatically holds that $$N_\epsilon \le \frac{C}{\epsilon^{-n}}$$, hence $$d=\dim_B M \le n$$.

By the definition of limit for any $$\delta>0$$ there exists $$\epsilon_0>0$$ such that for all $$\epsilon \in (0,\epsilon_0)$$ it holds $$d-\delta < \frac{\log N_\epsilon}{\log \epsilon^{-1}} < d+\delta$$, hence $$\epsilon^{-(d-\delta)} < N_\epsilon < \epsilon^{-(d+\delta)}$$. So if $$s > d+\delta$$ then $$N_\epsilon \epsilon^{s} \to 0$$ as $$\epsilon\to 0$$. Similarly, if $$0< s < d-\delta$$ then $$N_\epsilon \epsilon^{s} \to \infty$$ as $$\epsilon \to 0$$. This argument shows that $$s_0 \in [d-\delta, d+\delta]$$ and by arbitrariness of $$\delta>0$$ we conclude that $$s_0 = \delta$$.